How can I find general formula of this sequence? Let be the sequence $(a_n)$ so that $a_1 = 1$,  $a_{n+1}=\dfrac{20}{3 a_n+4}$, $\forall n \geqslant 1.$ I can prove that the sequence is an increasing sequence and convert to 2. But, I can find the general formula of this sequence. How can I find it?
 A: A useful idea is to look at the deviation from the equilibrium $a^*=2$, so set $b_n = a_n - 2$.
In terms of the sequence $(b_n)$, the recursion becomes
$$
b_{n+1} = \frac{-6 b_n}{3 b_n + 10}
,\qquad
b_1=-1
.
$$
Next set $c_n=1/b_n$. This gives a linear recurrence
$$
c_{n+1} = -\frac12 - \frac53 c_n
,\qquad
c_1=-1
.
$$
There is an equilibrium (unstable) at $c^* = -3/16$. So again look at the deviation from the equilibrium, by letting $d_n=c_n-(-3/16)$. This gives
$$
d_{n+1} = - \frac53 d_n
,\qquad
d_1=-\frac{13}{16}
.
$$
Can you take it from there?
A: $$A_{n+1}=\frac{20}{3A_n+4} \implies A_{n+1}~(3A_n +4)= 20.~~~(1)$$
Let $$(3A_n+4)=\frac{B_n}{B_{n-1}} \implies A_n=\frac{1}{3} \left(\frac{B_n}{B_{n-1}}-4\right)~~~(2)$$
Then (1) becomes $$B_{n+1}-60 B_{n-1}-4 B_n =0~~~(3)$$
Now let us put $B_n=t^n$ in (3) to get 
$$t^2-4t-60=0 \implies t=10,-6.$$
So the solution of (3) is $$B_n=P (10)^n +Q(-6)^n ~~~(4)$$
Inserting (4) in (2) we get $$A_n=\frac{1}{3} \left(\frac{6 R (10)^{n-1}-10 (-6)^{n-1}}{R (10)^{n-1}+ (-6)^{n-1}}\right), R=\frac{P}{Q}.$$
Using $A_1=1$ gives $$R=\frac{13}{3}.$$
Finally, we get $$A_n= \left( \frac{26~ (5)^{n-1}-10 ~ (-3)^{n-1}}{13~ (5)^{n-1} +3 ~(-3)^{n-1}} \right).$$
